# Solving a parallel RC circuit without knowing the capacitor resistance

I Have the following circuit scheme (I have to find the U): $$R_1=10, R_2=20 (\Omega); A_1->i=0.6, A_2->i=0.8(A)$$ I have a problem with calculating the impedance Z since I need it to calculate he U, but I don't know how to calculate it since Xc (capacitor resistance)**** is not given.

What I've done so far: $$i_1=i_2+i_3=1.4 A;\quad U_{r1}=i_1*R_1=14V;\quad U_{r2}=i_1*R_2=12V;$$ and after that I get stuck because I haven't got the Xl (inductor resistance) to solve for Ur3? How should I proceed?

I hope this question is appropriate for this community (my first question).

EDIT: Results that are given: • Presumably the point in your course teaches you that you can assume the amp-meters to represent zero resistance? Hopefully that's a decent hint. – Asmyldof Sep 11 '16 at 10:11
• But I wouldn't even take them into account in my problem, I just use the facet they give me the value of the current through that branch. I think I didn't quite understand... – eugene_sunic Sep 11 '16 at 10:25
• check my answers – saftargholi Sep 13 '16 at 14:00
• check my improved answer – Decapod Sep 13 '16 at 20:06

First remember current of capacitor and resistor not same, like this picture : • Uh what a complicated way. The point is we already have I1 as resistive and capacitive componets I1=I2+I3, then you find modulus, angle and then back to components. There's no need for this round back trip. Circuit can be solved in one single line. $U_1=U_\text{R2}+R_1(I_2+I_3)=R_2 I_2+ +R_1(I_2+I_3)= 20\,\Omega \times 0.6\,\text{A}+10\,\Omega\times (0.6+\text{j} 0.8\,\text{A})=18+\text{j}8\,\text{V}$ which in modulus yields $|U_1|=\sqrt{(18\,\text{V})^2+(8\,\text{V})^2}\approx 19.7\,\text{V}$ – carloc Sep 15 '16 at 6:32